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Oduvanchick [21]
3 years ago
11

A 1.2–liter tank at a temperature of 27°C contains a mixture of 1.5 moles of oxygen, 1.2 moles of nitrogen, and 1.6 moles of hyd

rogen. What is the total pressure of the mixture inside the tank? (R = 0.08205 L atm/K mol)
Chemistry
2 answers:
natka813 [3]3 years ago
7 0
You can use PV=nRT to figure this out, solving for P. In this case:

V = 1.2 L
n = 1.5 + 1.2 + 1.6 = 4.3 mol (assuming the question is giving moles of molecular O/N/H)
R = 0.08205 L atm/K mol
T = 27C = 300 K

P = nRT/V = 4.3 * 0.08205 * 300 / 1.2 = 88.2 atm
matrenka [14]3 years ago
7 0
Correct answer is 88.23 atmospheres.
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Here we apply the Clausius-Clapeyron equation:
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The atomic mass of an element is A. the sum of the protons and neutrons in one atom of the element. B. twice the number of proto
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Read 2 more answers
How many chlorine atoms would be in 6.02 X 10^23 units of gold III chloride
Pepsi [2]

Answer:

The number of chlorine atoms present in 6.02 \times 10^{23} units of gold III chloride is 18.066 \times 10^{23}

Explanation:

Formula of Gold (III) chloride: AuCl_{3}

<em>Avogadro Number</em> : Number of particles present in <u>one mole</u><u> </u>of a substance.

{N_{0}} =6.022 \times 10^{23}

Using,

n(moles)=\frac{Given\ number\ of\ particles}{N_{0}}

n =\frac{6.02\times 10^{23}}{6.022\times 10^{23}}

= 1 mole(0.9999 , nearly equal to 1 )

The given Gold III chloride sample is 1 mole in amount.

6.022 \times 10^{23}  = 1 mole of AuCl_{3}

In this Sample,

1 mole of AuCl_{3} will give = 3 mole of Chlorine atoms

1 mole of Cl contain = 6.022 \times 10^{23}

3 mole of Cl contain = 6.022 \times 10^{23}\times 3

3 mole of Cl contain =18.066 \times 10^{23}

So,

The number of chlorine atoms present in 6.02 \times 10^{23} units of gold III chloride is 18.066 \times 10^{23}

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4 years ago
What has fewer organisms then a domain but it also has more organisms then a phylum
kramer

Answer:

a species

Explanation:

6 0
3 years ago
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